洗牌后找出球的正确位置

考虑随机播放游戏。有3个编号从1到3的玻璃杯，其中一个球藏在任何一个玻璃杯的下面。然后，将任意两副眼镜洗净。进行此操作3次。

给定一个整数N，范围为[1、3]和3对相同范围的整数。第N个玻璃杯最初包含球，每对给定的整数表示需要重新排列的玻璃杯的索引。请记住，每次洗牌后，眼镜都会重新编号。

任务是在所有洗牌操作之后找出包含球的玻璃的索引。

例子：

Input:
N = 3
3 1
2 1
1 2
Output: 1
Firstly the 3rd glass contain the ball.
After the first shuffle operation (3, 1), 1st glass contain the ball.
After the second shuffle operation (2, 1), 2nd glass contain the ball.
After the third shuffle operation (1, 2), 1st glass contain the ball.

Input:
N = 1
1 3
1 2
2 3
Output: 2

方法：最简单的方法是为每个随机操作运行一个循环。

如果要洗牌的2个玻璃杯中的任何一个都包含球，则很明显要将N的值更改为要洗牌的玻璃的折射率。

如果2个混洗眼镜中的任何一个都不包含球，则无需执行任何操作。

下面是上述代码的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
const int M = 3, N = 2;
  
// Function to generate the index of the glass
// containing the ball
void getIndex(int n, int shuffle[][N])
{
    for (int i = 0; i < 3; i++) {
  
        // Checking if the glasses
        // being shuffled contain
        // the ball
  
        // Change the index
        if (shuffle[i][0] == n)
            n = shuffle[i][1];
  
        // Change the index
        else if (shuffle[i][1] == n)
            n = shuffle[i][0];
    }
  
    // Print the index
    cout << n;
}
  
// Driver's Code
int main()
{
    int n = 3;
  
    // Storing all the shuffle operation
    int shuffle[M][N] = {
        { 3, 1 },
        { 2, 1 },
        { 1, 2 }
    };
  
    getIndex(n, shuffle);
}

Java

// Java implementation of the above approach
import java.io.*;
  
class GFG
{
static int M = 3;
static int N = 2;
  
// Function to generate the index of the glass
// containing the ball
static void getIndex(int n, int shuffle[][])
{
    for (int i = 0; i < 3; i++) 
    {
  
        // Checking if the glasses
        // being shuffled contain
        // the ball
  
        // Change the index
        if (shuffle[i][0] == n)
            n = shuffle[i][1];
  
        // Change the index
        else if (shuffle[i][1] == n)
            n = shuffle[i][0];
    }
  
    // Print the index
    System.out.println (n);
}
  
// Driver Code
public static void main (String[] args) 
{
    int n = 3;
      
    // Storing all the shuffle operation
    int shuffle[][] = {{ 3, 1 },
                       { 2, 1 },
                       { 1, 2 }};
      
    getIndex(n, shuffle);
}
}
  
// This code is contributed by ajit.

Python3

# Python3 implementation of 
# the above approach 
M = 3; N = 2; 
  
# Function to generate the index of
# the glass containing the ball 
def getIndex(n, shuffle) :
  
    for i in range(3) :
  
        # Checking if the glasses 
        # being shuffled contain 
        # the ball 
  
        # Change the index 
        if (shuffle[i][0] == n) :
            n = shuffle[i][1]; 
  
        # Change the index 
        elif (shuffle[i][1] == n) :
            n = shuffle[i][0];
  
    # Print the index 
    print(n); 
  
# Driver Code 
if __name__ == "__main__" : 
  
    n = 3; 
  
    # Storing all the shuffle operation 
    shuffle = [[ 3, 1 ], 
               [ 2, 1 ], 
               [ 1, 2 ]]; 
  
    getIndex(n, shuffle); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
      
class GFG
{
static int M = 3;
static int N = 2;
  
// Function to generate the index of
// the glass containing the ball
static void getIndex(int n, int [,]shuffle)
{
    for (int i = 0; i < 3; i++) 
    {
  
        // Checking if the glasses
        // being shuffled contain
        // the ball
  
        // Change the index
        if (shuffle[i, 0] == n)
            n = shuffle[i, 1];
  
        // Change the index
        else if (shuffle[i, 1] == n)
            n = shuffle[i, 0];
    }
  
    // Print the index
    Console.WriteLine(n);
}
  
// Driver Code
public static void Main (String[] args) 
{
    int n = 3;
      
    // Storing all the shuffle operation
    int [,]shuffle = {{ 3, 1 },
                      { 2, 1 },
                       { 1, 2 }};
      
    getIndex(n, shuffle);
}
}
      
// This code is contributed by Princi Singh

输出：

1